Problem Statement
Points $A(1, 2, 0)$, $B(3, 0, 1)$, and $C(2, 4, 3)$ lie in a plane $\Pi$.
(a) Find the vectors $\vec{AB}$ and $\vec{AC}$. [2 Marks]
(b) Hence, find a normal vector $\vec{n}$ to the plane $\Pi$. [3 Marks]
(c) State the Cartesian equation of plane $\Pi$ in the form $ax + by + cz = d$. [2 Marks]
Verified Solution & Marking Scheme
Part (a): Compute Direction Vectors
$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}$
$\vec{AC} = \vec{OC} - \vec{OA} = \begin{pmatrix} 2 \\ 4 \\ 3 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$
Part (b): Vector Cross Product
$\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & -2 & 1 \\ 1 & 2 & 3 \end{vmatrix}$
$= \vec{i}((-2)(3) - (1)(2)) - \vec{j}((2)(3) - (1)(1)) + \vec{k}((2)(2) - (-2)(1))$
$= \vec{i}(-6 - 2) - \vec{j}(6 - 1) + \vec{k}(4 + 2) = -8\vec{i} - 5\vec{j} + 6\vec{k}$
We can choose $\vec{n} = \begin{pmatrix} -8 \\ -5 \\ 6 \end{pmatrix}$ or $\begin{pmatrix} 8 \\ 5 \\ -6 \end{pmatrix}$.
Part (c): Cartesian Plane Equation
Using $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$ with $A(1, 2, 0)$:
$8x + 5y - 6z = 8(1) + 5(2) - 6(0) = 8 + 10 = 18$
So the plane equation is $8x + 5y - 6z = 18$ (or $-8x - 5y + 6z = -18$).